Solid Mensuration 1 2.docx

Solid Mensuration Circle = It is a closed figure with set of points that is equidistant to a fixed point called center. 𝝅𝒅𝟐 𝑨𝜟 = 𝝅𝒓𝟐 = 𝟒 𝑪𝑶 = 𝟐𝝅𝒓 Example: Solve for the area and circumference of a circle whose radius is 10cm. 𝑨𝜟 = 𝜋𝑟 2 = 𝜋(10)2 = 𝟑𝟏𝟒. 𝟏𝟔𝒄𝒎𝟐 𝐶𝑂 = 2𝜋𝑟 = 2𝜋(10) = 𝟔𝟐. 𝟖𝟑𝒄𝒎 Arc Length = It is a set of definite points that are equidistant to a fixed point called center. 𝑺𝒂𝒓𝒄 =

𝝅𝜽𝒓 𝟏𝟖𝟎

Sector = A closed figure bounded by an arc and two radii. 𝑨𝜟 = 𝑷=

𝝅𝜽𝒓𝟐 𝟑𝟔𝟎

𝝅𝜽𝒓 𝝅𝜽 + 𝟐𝒓 = 𝒓 (𝟐 + ) 𝟏𝟖𝟎 𝟏𝟖𝟎

Example: Solve for the area and perimeter of a sector whose radius is 10cm and its central angle is 36o. 𝑨𝜟 =

𝜋𝜃𝑟 2 𝜋(36)(10)2 = = 𝟑𝟏. 𝟒𝟐𝒄𝒎𝟐 360 360

𝑃 = 𝑟 (2 +

𝜋𝜃 36𝜋 ) = 10 (2 + ) = 𝟐𝟔. 𝟐𝟖𝒄𝒎 180 180

Segment = A closed figure bounded by an arc and a chord. 𝝅𝜽𝒓𝟐 𝒓𝟐 𝑺𝒊𝒏𝜽 𝝅𝜽 𝑺𝒊𝒏𝜽 𝑨𝜟 = − = 𝒓𝟐 ( − ) 𝟑𝟔𝟎 𝟐 𝟑𝟔𝟎 𝟐 Example: Solve for the area of a segment whose radius is 10cm and its central angle is 36o. 𝑨𝜟 = 𝒓𝟐 (

𝝅𝜽 𝑺𝒊𝒏𝜽 𝟑𝟔𝝅 𝑺𝒊𝒏𝟑𝟔 − ) = 𝟏𝟎𝟐 ( − ) = 𝟐. 𝟎𝟑𝒄𝒎𝟐 𝟑𝟔𝟎 𝟐 𝟑𝟔𝟎 𝟐

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Area of Triangle Area = It is the surface included within the set of lines. Perimeter = it is the boundary of a closed plane figure.

h a

c

a

h

c

a

c

b

b b 𝒃𝒉 𝑨𝜟 = 𝟐 𝑷= 𝒂+𝒃+𝒄 Example: Solve the area and perimeter of a right triangle whose base is 15cm and its altitude is 5cm. 𝒃𝒉 𝟓(𝟏𝟓) 𝑨= = = 𝟑𝟕. 𝟓𝒄𝒎𝟐 𝟐 𝟐 𝑐 = √52 + 152 = √25 + 225 = √250 = 𝟏𝟓. 𝟖𝟏𝒄𝒎 𝑃 = 𝑎 + 𝑏 + 𝑐 = 5 + 15 + 15.81 = 𝟑𝟓. 𝟖𝟏𝒄𝒎 C

b

a A B c 1. Heron’s Formula: 𝟐

𝑨𝜟 = √𝒔(𝒔 − 𝒂)(𝒔 − 𝒃)(𝒔 − 𝒄) Where: AΔ = Area of the triangle a, b, c = sides of the oblique triangle 𝒂+𝒃+𝒄 𝟐 Example: The coordinates of an oblique triangle are (4, 5), (-2, 1) and (6, -4), what is the area and perimeter of the triangle? A(4, 5) B(-2, 1) C(6, -4) 𝑃 = 𝑎 + 𝑏 + 𝑐 = 7.21 + 9.43 + 9.22 = 𝟐𝟓. 𝟖𝟔𝒄𝒎 2 2 2 2 𝐴𝐵 = √(4 − −2) + (5 − 1) = √6 + 4 = √52 = 𝟕. 𝟐𝟏 𝑃 25.86 𝑠= = = 𝟏𝟐. 𝟗𝟑 𝐵𝐶 = √(−2 − 6)2 + (1 − −4)2 = √82 + 52 2 2 = √89 = 𝟗. 𝟒𝟑 2 𝐶𝐴 = √(6 − 4)2 + (−4 − 5)2 = √22 + 92 𝑨𝜟 = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐) = √85 = 𝟗. 𝟐𝟐 = √12.93(5.72)(3.5)(3.71) = 𝟑𝟎. 𝟗𝟗𝒔𝒖 𝒔=

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2. Coordinate Method: 𝑨𝜟 =

𝟏 𝒙𝟏 [ 𝟐 𝒚𝟏

𝒙𝟐 𝒙𝟑 𝒚𝟐 𝒚𝟑

𝒙𝟏 𝒚𝟏 ]

Example: Solve the previous example by coordinate method. 𝑨𝜟 =

𝟏 𝟒 −𝟐 𝟔 𝟒 𝟏 [ ] = ((𝟒 + 𝟖 + 𝟑𝟎)— (−𝟏𝟎 + 𝟔 − 𝟏𝟔)) 𝟓 𝟏 −𝟒 𝟓 𝟐 𝟐 =

𝟏 ((𝟒𝟐)— (−𝟐𝟎)) = 𝟑𝟏𝒔𝒖 𝟐

3. Two sides and the included angle known 𝑨𝜟 =

𝒂𝒄𝑺𝒊𝒏 𝑩 𝟐

Example: In triangle ABC, a = 40m, c = 50m and angle A = 53o. Find the area of the triangle? 𝑆𝑖𝑛𝐶 𝑆𝑖𝑛53 = 50 40 50𝑆𝑖𝑛53 𝐶1 = 𝑆𝑖𝑛−1 ( ) = 86.65 40 𝐶2 = 180 − 86.65 = 93.35 𝐵1 = 180 − 86.65 − 53 = 40.35 𝐵2 = 180 − 93.35 − 53 = 33.65 𝑨𝜟 =

𝑎𝑐𝑆𝑖𝑛 𝐵 40(50)𝑆𝑖𝑛40.35 = = 𝟔𝟒𝟕. 𝟒𝟔𝒎𝟐 2 2

𝑨𝜟 =

𝑎𝑐𝑆𝑖𝑛 𝐵 40(50)𝑆𝑖𝑛33.65 = = 𝟓𝟓𝟒. 𝟏𝟐𝒎𝟐 2 2

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4. One side and two adjacent angles known 𝒃𝟐 𝑺𝒊𝒏𝑨 𝑺𝒊𝒏 𝑪 𝑨𝜟 = 𝟐𝑺𝒊𝒏 𝑩 Example: The area of a triangle is 8346m2 and two of its interior angles are 37o25’ and 56o17’. What is the length of the longest side? 𝐵 = 180 − 37°25′ − 56°17′ = 𝟖𝟔. 𝟑 𝑨𝜟 =

𝒃𝟐 𝑺𝒊𝒏𝑨 𝑺𝒊𝒏 𝑪 𝟐𝑺𝒊𝒏 𝑩

8346 = 𝑏2 =

𝑏 2 𝑆𝑖𝑛37°25′ 𝑆𝑖𝑛 56°17′ 2𝑆𝑖𝑛 86.3

8346(2)𝑆𝑖𝑛 86.3 𝑆𝑖𝑛37°25′ 𝑆𝑖𝑛 56°17′

8346(2)𝑆𝑖𝑛 86.3 𝑏=√ = 181.54𝑚 𝑆𝑖𝑛37°25′ 𝑆𝑖𝑛 56°17′ 5. Triangle inscribe in a circle 𝑨𝜟 =

𝒂𝒃𝒄 𝟒𝒓

Example: The sides of a triangle are 8cm, 10cm and 14cm respectively. Determine the radius of the circle if the triangle is inscribed in a circle. 𝑠=

8+10+14 2

=

32 2

= 𝟏𝟔

2

𝐴𝛥 = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐) 2

√16(16 − 8)(16 − 10)(16 − 14)

= √16(8)(6)(2) = 𝟑𝟗. 𝟏𝟗𝒄𝒎𝟐 𝑨𝜟 =

𝒂𝒃𝒄 𝟒𝒓

39.19 =

𝑟=

8(10)(14) 4𝑟

8(10)(14) = 7.14 4(39.19)

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6. The circle is inscribed in a triangle 𝑨 = 𝒓𝒔 Example: The sides of the triangle are 8, 10 and 14 cm. Find the radius of the circle inscribed in the triangle. 𝑠=

8 + 10 + 14 32 = = 𝟏𝟔 2 2

𝐴𝛥 = 39.19𝑐𝑚2 𝐴𝛥 = 𝑟𝑠 39.19 = 𝑟(16) 𝑟=

39.19 = 16

2.45cm

7. Triangle with an escribed circle 𝑨 = 𝒓(𝑺 − 𝒂) Example: The radius of the circle escribed outside a triangle is 12cm and is tangent to the 18cm side. The other side of the triangle is 24cm long. If the area of the triangle is 216 sq.cm., find the third side of the triangle. 𝑨 = 𝒓(𝑺 − 𝒂) 𝟐𝟏𝟔 = 𝟏𝟐(𝑺 − 𝟏𝟖) 𝑠=

216 + 18 = 𝟑𝟔 12

𝑠=

𝑎+𝑏+𝑐 2

36 =

18 + 24 + 𝑐 2

𝑐 = 36(2) − 18 − 24 = 𝟑𝟎

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8. Length of the median of a triangle 𝒉𝒄 =

𝟏𝟐 √𝟐𝒂𝟐 + 𝟐𝒃𝟐 − 𝒄𝟐 𝟐

Example: If the sides of the triangle ABC are AB = 15cm, BC = 18cm and CA = 24cm, compute the length of the median from C to the side AB ℎ𝑐 =

12 √2𝑎2 + 2𝑏 2 − 𝑐 2 2

ℎ𝑐 =

12 √2(18)2 + 2(24)2 − (15)2 = 19.84𝑐𝑚 2

9. Length of bisector of an angle of a triangle to opposite side 𝟐√𝒃𝒄𝒔(𝒔 − 𝒂) 𝒉𝒄 = 𝒃+𝒄 Example: Triangle ABC has sides AB = 14cm, BC = 18cm and CA = 24cm respectively. Find the length of the bisector of angle A to the side BC. 𝑠=

𝑎+𝑏+𝑐 2

𝑠=

14 + 24 + 18 = 28𝑐𝑚𝑠 2

ℎ𝑐 =

2√𝑏𝑐𝑠(𝑠 − 𝑎) 𝑏+𝑐

ℎ𝑐 =

2√24(14)(28)(28 − 18) = 16.14𝑐𝑚 24 + 14

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10. Polygon Inscribed in a circle 𝟑𝟔𝟎 𝒏 Where: φ = central angle of an isosceles triangle formed from a polygon n = number of sides of a polygon ∅=

𝟏𝟖𝟎 − ∅ 𝟐 Where: 𝛽 = interior angle of an isosceles triangle formed from a polygon 𝜷=

𝑨 = 𝒏𝒓𝟐 𝑺𝒊𝒏𝜷𝑪𝒐𝒔𝜷 𝑷 = 𝟐𝒏𝒓𝑪𝒐𝒔𝜷 Example: Compute the area and perimeter of a pentagon inscribed in a circle, r =10cm. ∅=

360 360 = = 𝟕𝟐 𝑛 5

𝛽=

180 − ∅ 180 − 72 = = 𝟓𝟒 2 2

𝐴 = 𝑛𝑟 2 𝑆𝑖𝑛𝛽𝐶𝑜𝑠𝛽 = 5(10)2 𝑆𝑖𝑛54𝐶𝑜𝑠54 = 𝟐𝟑𝟕. 𝟕𝟔𝒄𝒎𝟐 𝑃 = 2𝑛𝑟𝐶𝑜𝑠𝛽 = 2(5)(10)𝐶𝑜𝑠54 = 𝟓𝟖. 𝟕𝟖𝒄𝒎 11. Polygon Circumscribing a Circle 𝒏𝒓𝟐 𝑻𝒂𝒏𝜷 𝟐𝒏𝒓 𝑷= 𝑻𝒂𝒏𝜷 𝑨=

Example: Compute the area and perimeter of a pentagon circumscribing a circle, r = 10cm. ∅=

360 360 = = 𝟕𝟐 𝑛 5

𝛽=

180 − ∅ 180 − 72 = = 𝟓𝟒 2 2

𝑃=

2𝑛𝑟 2(5)(10) = = 𝟕𝟐. 𝟔𝟓𝒄𝒎 𝑇𝑎𝑛𝛽 𝑇𝑎𝑛54

𝐴=

𝑛𝑟 2 5(10)2 = = 𝟑𝟔𝟑. 𝟐𝟕𝒄𝒎𝟐 𝑇𝑎𝑛𝛽 𝑇𝑎𝑛54

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Name: Course/Year & Section: Compute the area and perimeter of the following if r = 15cm. 1.a. Inscribed Octagon ∅=

360 360 = = 𝟒𝟓 𝑛 8

𝛽=

180 − ∅ 180 − 72 = = 𝟔𝟕. 𝟓 2 2

𝐴 = 𝑛𝑟 2 𝑆𝑖𝑛𝛽𝐶𝑜𝑠𝛽 = 8(15)2 𝑆𝑖𝑛67.5𝐶𝑜𝑠67.5 = 𝟔𝟑𝟔. 𝟒𝟎𝒄𝒎𝟐 𝑃 = 2𝑛𝑟𝐶𝑜𝑠𝛽 = 2(8)(15)𝐶𝑜𝑠67.5 = 𝟗𝟏. 𝟖𝟓𝒄𝒎 1.b. Circumscribing Octagon 𝑃=

2𝑛𝑟 2(8)(15) = = 𝟗𝟗. 𝟒𝟏𝒄𝒎 𝑇𝑎𝑛𝛽 𝑇𝑎𝑛67.5

𝐴=

𝑛𝑟 2 8(15)2 = = 𝟕𝟒𝟓. 𝟓𝟖𝒄𝒎𝟐 𝑇𝑎𝑛𝛽 𝑇𝑎𝑛67.5

2.a. Inscribed Decagon ∅=

360 360 = = 𝟑𝟔 𝑛 10

𝛽=

180 − ∅ 180 − 36 = = 𝟕𝟐 2 2

𝐴 = 𝑛𝑟 2 𝑆𝑖𝑛𝛽𝐶𝑜𝑠𝛽 = 10(15)2 𝑆𝑖𝑛72𝐶𝑜𝑠72 = 𝟔𝟔𝟏. 𝟐𝟔𝒄𝒎𝟐 𝑃 = 2𝑛𝑟𝐶𝑜𝑠𝛽 = 2(10)(15)𝐶𝑜𝑠72 = 𝟗𝟐. 𝟕𝟏𝒄𝒎 2.b. Circumscribing Decagon 𝑃=

2𝑛𝑟 2(10)(15) = = 𝟗𝟕. 𝟒𝟖𝒄𝒎 𝑇𝑎𝑛𝛽 𝑇𝑎𝑛72

𝐴=

𝑛𝑟 2 10(15)2 = = 𝟕𝟑𝟏. 𝟎𝟕𝒄𝒎𝟐 𝑇𝑎𝑛𝛽 𝑇𝑎𝑛72

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Ellipse 𝑨 = 𝝅𝒂𝒃 (𝒂𝟐 + 𝒃𝟐 ) 𝑷 = 𝟐𝝅√ 𝟐 Example: a = 5cm, b = 4cm 𝐴 = 𝜋𝑎𝑏 = 𝜋(5)(4) = 20𝜋 = 𝟔𝟐. 𝟖𝟑𝒄𝒎𝟐 (𝑎2 + 𝑏 2 ) (52 + 42 ) 25 + 16 41 𝑃 = 2𝜋√ = 2𝜋√ = 2𝜋√ = 2𝜋√ = 𝟐𝟖. 𝟒𝟓𝒄𝒎 2 2 2 2 Spandrel = it is sometimes referred as the ornamented space between the left or right exterior curve of a line and the enclosing right triangle. Line Horizontal Diagonal Curve 𝑨=

n 0 1 2 or >

h

𝒃𝒉 𝒏+𝟏

n = degree of the curve

b

Example: Determine the area of a second degree curve whose base is 10cm and height 5cm. 𝑏ℎ 10(5) 50 𝐴= = = = 𝟏𝟔. 𝟔𝟕𝒄𝒎𝟐 𝑛+1 2+1 3

Area of Quadrilaterals Square = A quadrilateral whose sides are equal and consecutive sides are perpendicular to each other 𝑨 = 𝒔𝟐 s P = 4s s Example: Solve the area and perimeter of a square whose side is 5cm each. 𝑨 = 𝒔𝟐 = 𝟓𝟐 = 𝟐𝟓𝒄𝒎𝟐 P = 4s = 4(5) = 20cm Rectangle = A quadrilateral whose opposite side are equal in length and consecutive sides are perpendicular to each other.

h

A = bh P = 2(b + h)

b Example: Solve the area and perimeter of a rectangle whose height is 5cm and base is 10cm. A = bh = 5(10) = 50cm2 Page 9 of 25

P = 2(b + h) = 2(5+10) = 30cm Rhombus = A quadrilateral whose sides are all equal but the consecutive sides are not perpendicular to each other. 𝐴 = 𝑠ℎ = 𝑠 2 𝑆𝑖𝑛∅ P =4s

s h φ

s Example: 1. Solve the area and perimeter of a rhomboid whose sides are 10cm and is inclined at 30o. 𝐴 = 𝑠ℎ = 𝑠 2 𝑆𝑖𝑛∅ = 102 𝑆𝑖𝑛30 = 100(0.5) = 𝟓𝟎𝒄𝒎𝟐 P =4s = 4(10) = 40cm 2. The area of a rhombus is 168sq.m. If one of its diagonal is 12m, find the length of the sides of a rhombus.

𝒅𝟏 𝒅𝟐 𝟐 12𝑑2 168 = 2 𝑑2 = 28𝑐𝑚 𝑨=

𝑥 = √62 + 142 = 𝟏𝟓. 𝟐𝟑𝒄𝒎 Rhomboid= A quadrilateral whose opposite sides are equal in length but the consecutive sides are not perpendicular to each other.

s

h

𝑨 = 𝒃𝒉 = 𝒃𝒔𝑺𝒊𝒏∅ P = 2(s + b)

φ b Example: 1. Solve the area and perimeter of a rhomboid whose base is 10cm and slant height is 5cm which is inclined at 30o. 𝐴 = 𝑏𝑠𝑆𝑖𝑛∅ = 10(5)𝑆𝑖𝑛30 = 𝟐𝟓𝒄𝒎𝟐 P = 2(s + b) = 2(5 + 10) = 30cm

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2. The diagonals of a parallelogram are 18cm and 30cm respectively. One side of the parallelogram is 12cm. Find the area of the parallelogram. 15 9

θ

12 𝜃 = 𝐶𝑜𝑠 −1 (

122 − 152 − 92 ) = 53.13 −2(15)(9)

𝒅𝟏 𝒅𝟐 𝑺𝒊𝒏𝜽 𝟐 18(30)𝑆𝑖𝑛53.13 𝐴= = 𝟐𝟏𝟔𝒄𝒎𝟐 2 Trapezoid = A quadrilateral whose opposite bases are parallel. a 𝑨=

𝑨=𝒉 c

h

d

(𝒂+𝒃) 𝟐

= 𝒎𝒉 =

𝒃𝟐 − 𝒂𝟐 𝟐(𝑪𝒐𝒕𝜽 + 𝑪𝒐𝒕𝜷)

=

(𝒃𝟐 − 𝒂𝟐 )𝑻𝒂𝒏𝜽𝑻𝒂𝒏𝜷 𝟐(𝑻𝒂𝒏𝜽+𝑻𝒂𝒏𝜷)

P=a+b+c+d

m θ

β

b Example: 1. Find the area and perimeter of trapezoid having length of median equal to 32m and an altitude of 6m. 𝐴 = 𝑚ℎ = 32(6) = 𝟏𝟗𝟐𝒎𝟐 2. If a=100m, b=250m, θ=75o, β=60o, find the area and perimeter of the trapezoidal lot shown. 100 x

y

75 150 60

100

(𝑏 2 − 𝑎2 )𝑇𝑎𝑛𝜃𝑇𝑎𝑛𝛽 (2502 − 1002 )𝑇𝑎𝑛75𝑇𝑎𝑛60 = = 𝟑𝟏, 𝟎𝟓𝟒. 𝟎𝟖𝒎𝟐 2(𝑇𝑎𝑛𝜃 + 𝑇𝑎𝑛𝛽) 2(𝑇𝑎𝑛75 + 𝑇𝑎𝑛60) 𝑥 150 𝑦 150 = = 𝑆𝑖𝑛60 𝑆𝑖𝑛45 𝑆𝑖𝑛75 𝑆𝑖𝑛45 𝐴=

x= 183.71

y= 204.90

P = a + b + c + d = 100+250+183.71+204.90 = 738.61m Four sides not parallel having sides AB = 10cm, BC = 5cm, CD = 14.14cm and DA = 15cm, if the sum of the opposite angles is equal to 225o. 𝑨 = √(𝒔 − 𝒂)(𝒔 − 𝒃)(𝒔 − 𝒄)(𝒔 − 𝒅) − 𝒂𝒃𝒄𝒅𝑪𝒐𝒔𝟐 𝜽

Page 11 of 25

𝒔=

𝒂+𝒃+𝒄+𝒅 𝟐

Where: θ = average of opposite angles Examples: Find the area of a quadrilateral having sides AB=10cm, BC=5cm, CD=14.14cm and DA=15cm, if the sum of the opposite angles is equal to 225o. 𝑎+𝑏+𝑐+𝑑 10 + 5 + 14.14 + 15 𝑠= = = 22.07 2 2 𝜃=

225 = 112.5 2

𝐴 = √(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)(𝑠 − 𝑑) − 𝑎𝑏𝑐𝑑𝐶𝑜𝑠 2 𝜃 𝐴 = √(22.07 − 10)(22.07 − 5)(22.07 − 14.14)(22.07 − 15) − 10(5)(14.14)(15)𝐶𝑜𝑠 2 112.5 𝐴 = √(12.07)(17.07)(7.93)(7.07) − 10(5)(14.14)(15)𝐶𝑜𝑠 2 112.5 𝑨 = 𝟗𝟗. 𝟗𝟗𝒄𝒎𝟐 Quadrilateral Inscribed in a Circle 𝑨 = √(𝒔 − 𝒂)(𝒔 − 𝒃)(𝒔 − 𝒄)(𝒔 − 𝒅) Ptolemy’s Theorem 𝑨𝑩(𝑫𝑪) + 𝑨𝑫(𝑩𝑪) = 𝑨𝑪(𝑩𝑫) Example: Find the area of quadrilateral ABCD inscribed in a circle if AB=90m, DA=50m, CD=70m, BD=101.76m and AC=97.29m.

A

B

𝐴𝐵(𝐷𝐶) + 𝐴𝐷(𝐵𝐶) = 𝐴𝐶(𝐵𝐷) 90(70) + 50(𝐵𝐶) = 97.29(101.76) 𝐵𝐶 = 72𝑚 𝑠=

𝑎+𝑏+𝑐+𝑑 90 + 70 + 72 + 50 = = 141 2 2

C D

𝐴 = √(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)(𝑠 − 𝑑) 𝐴 = √(141 − 90)(141 − 70)(141 − 72)(141 − 50) 𝑨 = 𝟒𝟕𝟔𝟖. 𝟐𝟔𝒎𝟐

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Cyclic Quadrilateral Circumscribing a Circle 𝑨 = √𝒂𝒃𝒄𝒅 Example: Find the area of a cyclic quadrilateral circumscribing a circle having sides of 48m, 32m, 54m and 38m respectively. 𝐴 = √𝑎𝑏𝑐𝑑 = √48(32)(54)(38) = 𝟏𝟕𝟕𝟓. 𝟑𝟓𝒎𝟐

Volume = It is the amount of space occupied by a three dimensional object as measured in cubic units. 1. Cube = It is a regular solid of six equal squares sides. 𝑽 = 𝒔𝟑 𝑨𝑻 = 𝟔𝒔𝟐

s s

𝟐

𝑨𝑳 = 𝟒𝒔

s

Example: Solve for the volume and lateral area of the cube whose side is 10cm. 𝑉 = 𝑠 3 = 103 = 𝟏𝟎𝟎𝟎𝒄𝒎𝟑 𝐴 𝑇 = 6𝑠 2 = 6(10)2 = 6(100) = 𝟔𝟎𝟎𝒄𝒎𝟐 𝐴𝐿 = 4𝑠 2 = 4(10)2 = 4(100) = 𝟒𝟎𝟎𝒄𝒎𝟐

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2. Prism = It is a polyhedron with two polygonal faces lying in parallel planes and with the other faces parallelograms. 𝑽 = 𝒃𝒘𝒉 𝑨𝑻 = 𝟐(𝒃𝒉 + 𝒃𝒘 + 𝒉𝒘) 𝑽 = 𝑨𝑩 𝒉

Example: a. Solve for the volume and lateral area of the prism whose base is 10cm, width 15cm and a height of 5cm. 𝑉 = 𝑏𝑤ℎ = 10(15)(5) = 𝟕𝟓𝟎𝒄𝒎

𝟑

𝐴𝐿 = 2(𝑏ℎ + ℎ𝑤) = 2(10(5) + 5(15)) = 𝟐𝟓𝟎𝒄𝒎𝟐 𝐴 𝑇 = 2(𝑏ℎ + 𝑏𝑤 + ℎ𝑤) = 2(10(5) + 10(15) + 5(15)) = 𝟓𝟓𝟎𝒄𝒎𝟐

b. Solve for the volume and lateral area of the prism whose base area is an equilateral triangle with side of 10cm and a height of 5cm 𝑉=

𝑎𝑐𝑆𝑖𝑛𝜃ℎ 102 𝑆𝑖𝑛60(5) = = 𝟐𝟏𝟔. 𝟓𝟏𝒄𝒎𝟑 2 2

𝐴𝐿 = 3(10)(5) = 𝟏𝟓𝟎 𝒄𝒎𝟐 𝐴 𝑇 = 150 + 2 (

102 𝑆𝑖𝑛60 ) = 𝟐𝟑𝟔. 𝟔 𝒄𝒎𝟐 2

3. Cylinder = It is a prism whose base area is a circle 𝑽 = 𝝅𝒓𝟐 𝒉 𝑨𝑳 = 𝟐𝝅𝒓𝒉 𝑨𝑻 = 𝟐𝝅𝒓𝒉 + 𝟐𝝅𝒓𝟐 = 𝟐𝝅𝒓(𝒓 + 𝒉)

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Example: Solve for the volume and total area of the cylinder whose radius is 10cm and altitude of 5cm. 𝑉 = 𝜋𝑟 2 ℎ = 𝜋(10)2 (5) = 𝟏𝟓𝟕𝟎. 𝟖𝒄𝒎𝟑 𝐴𝐿 = 2𝜋𝑟ℎ = 2𝜋(10)(5) = 𝟗𝟒𝟐. 𝟒𝟖𝒄𝒎𝟐 𝐴 𝑇 = 2𝜋𝑟(𝑟 + ℎ) = 2𝜋(10)(10 + 5) = 𝟗𝟒𝟐. 𝟒𝟖𝒄𝒎𝟐

4. Pyramid = It is a polyhedron whose base is a polygon and triangular faces with a common vertex. 𝑽=

𝒃𝒘𝒉 𝑨𝑩 𝒉 = 𝟑 𝟑

For square base 𝑨𝑳 = 𝟐𝒔𝒍 𝑨𝑻 = 𝟐𝒔𝒍 + 𝒔𝟐 For rectangular base 𝑨𝑳 = 𝒃𝒍𝟏 + 𝒘𝒍𝟐 𝑨𝑻 = 𝒃𝒍𝟏 + 𝒘𝒍𝟐 + 𝒃𝒘 For triangular base 𝟑𝒔𝒍 𝑨𝑳 = 𝟐 𝟑𝒔𝒍 𝒔𝟐 𝑺𝒊𝒏𝟔𝟎 𝑨𝑻 = + 𝟐 𝟐 Example: Solve for the volume and total area of the pyramid whose base area is square with side of 10cm and altitude of 5cm. 𝑏𝑤ℎ 𝐴𝐵 ℎ 10(10)(5) 𝑉= = = = 𝟏𝟔𝟔. 𝟔𝟕𝒄𝒎𝟑 3 3 3 𝐴𝐿 = 2𝑠𝑙 = 2(10)(5√2 ) = 𝟏𝟒𝟏. 𝟒𝟐𝒄𝒎𝟐 𝐴 𝑇 = 2𝑠𝑙 + 𝑠 2 = 141.42 + 102 = 𝟐𝟒𝟏. 𝟒𝟐𝒄𝒎𝟐

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5. Cone = It is a solid bounded by a circular plane base and the surface formed by line segments joining every point of the boundary of the base to a common vertex.

𝐴𝐿 = 𝜋𝑟𝐿

Example: a. The lateral area of a right circular cone is 47.12sq.cm. Find the volume of the cone if it has a slant height of 5cm. 𝐴𝐿 = 𝜋𝑟𝐿 47.12 = 𝜋𝑟(5) 𝑟 = 3𝑐𝑚 ℎ = √𝐿2 − 𝑟 2 = √52 − 32 = 4𝑐𝑚 𝑉=

𝜋𝑟 2 ℎ 𝜋(3)2 (4) = = 𝟑𝟕. 𝟕𝒄𝒎𝟑 3 3

Relationship of volume and height 𝑽𝟏 𝒉𝟏 = 𝑽𝑻 𝒉𝑻 Example: a. Given the volume of water in a cone is only ½ of that of the cone. If the height of water is 6cm, what is the height of the cone? 𝑉𝑇 𝑉1 = 2 𝑉1 63 = 3 𝑉𝑇 ℎ𝑐 𝑉𝑇 3 2 = 6 𝑉𝑇 ℎ𝑐 3 1 63 = 3 2 ℎ𝑐 𝒉𝒄 = 𝟕. 𝟓𝟔𝒄𝒎

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b. A circular cone having an altitude of 9m is divided into 2 segments having the same vertex. If the smaller altitude is 6m, find the ratio of the volume of small cone to the big cone. 𝑉1 ℎ1 3 = 𝑉𝑇 ℎ𝑇 3 𝑉1 63 = 3 = 𝟎. 𝟑𝟎 𝑉𝑇 9

Cone constructed from a sector 𝜽𝑹 𝑺 = 𝑹𝜽, 𝒓= 𝟑𝟔𝟎 Where: S = arc length of the sector or the circumference of the circle of the cone formed from the sector R = radius of the sector r = radius of the circle base of the cone θ = central angle of the sector Example: a. Find the volume of a cone to be constructed from a sector having a diameter of 72cm and a central angle of 210o. 𝜃𝜋𝑟 210𝜋(36) 𝑆= = = 131.95 180 180 𝑟=

𝜃𝑅 210(36) = = 21𝑐𝑚 360 360

ℎ = √362 − 212 ℎ = 29.24𝑐𝑚 𝑉=

𝐴𝐵 ℎ 131.95(29.24) = = 𝟏𝟐𝟖𝟔. 𝟎𝟕𝒄𝒎𝟐 3 3

b. The axis of the cone makes an angle of 60o with the horizontal. If the length of the axis is 30cm and its base radius 20cm, compute for the volume of the curve.

30

h

60

Page 17 of 25

ℎ 30 ℎ = 25.98𝑐𝑚 𝑆𝑖𝑛60 =

𝑉=

𝜋𝑟 2 ℎ 𝜋(20)2 (25.98) = = 𝟏𝟎, 𝟖𝟖𝟐. 𝟒𝟖𝒄𝒎𝟑 3 3

6. Sphere = It is a solid bounded by a surface consisting of all points at a given distance from a point constituting its center. 𝟒𝝅𝒓𝟑 𝟑 𝑨𝑳 = 𝟒𝝅𝒓𝟐 𝑽=

Example: What is the lateral area and volume of a sphere whose radius is 10cm? 𝐴𝐿 = 4𝜋𝑟 2 = 4𝜋(10)2 = 𝟏𝟐𝟓𝟔. 𝟔𝟒𝒄𝒎𝟐 𝑉=

4𝜋𝑟 3 4𝜋(10)3 = = 𝟒𝟏𝟖𝟖. 𝟕𝟗𝒄𝒎𝟑 3 3

Spherical Wedge and Spherical Lune 𝝅𝜽𝒓𝟑 𝑽𝑺𝑾 = 𝟐𝟕𝟎 𝝅𝜽𝒓𝟐 𝑨𝑺𝑳 = 𝟗𝟎 Example: a. The area of a lune is 30sq.m. If the area of a sphere is 120sq.m., what is the angle of the lune? 120 = 4𝜋𝑟 2 𝑟 = 3.09𝑚 𝜋𝜃𝑟 2 𝐴𝑆𝐿 = 90 𝜋𝜃(3.09)2 30 = 90 𝜽 = 𝟗𝟎. 𝟎𝟏 b. Find the volume and area of a spherical wedge whose central angle is π/5 radians and a radius of 6cm.

Page 18 of 25

𝑉𝑆𝑊 =

𝜋𝜃𝑟 3 𝜋(36)(6)3 = = 𝟗𝟎. 𝟒𝟖𝒄𝒎𝟑 270 270

𝐴𝑆𝐿 =

𝜋𝜃𝑟 2 𝜋(36)(6)2 = = 𝟒𝟓. 𝟐𝟒𝒄𝒎𝟑 90 90

Spherical Segment 𝝅𝒉𝟐 (𝟑𝒓 − 𝒉) 𝑽= 𝟑 𝑽=

𝝅𝒉[𝟑(𝒓𝟐𝟏 + 𝒓𝟐𝟐 ) + 𝒉𝟐 ] 𝟔

𝑨𝒁 = 𝟐𝝅𝒓𝒉 Example: a. The area of a zone of a spherical segment is 180π sq.m. If the radius of the sphere is 15m, compute the volume of the spherical segment. 𝐴𝑍 = 2𝜋𝑟ℎ 180𝜋 = 2𝜋(15)ℎ ℎ = 6𝑚 𝑉=

𝜋ℎ2 (3𝑟 − ℎ) 𝜋(6)2 (3(15) − 6) = = 𝟏𝟒𝟕𝟎. 𝟐𝟕𝒄𝒎𝟑 3 3

b. Find the volume of a spherical segment the radii of whose bases are 4m and 5m respectively with an altitude of 6m. 𝑉=

𝜋ℎ[3(𝑟12 + 𝑟22 ) + ℎ2 ] 𝜋(6)[3(42 + 52 ) + 62 ] = = 𝟒𝟗𝟗. 𝟓𝟏𝒄𝒎𝟑 6 6

c. A mixture compound from equal parts of two liquids, one white and the other black, was placed in a hemispherical bowl. The total depth of the two liquids is 6 inches. After standing for a short time the mixture separated, the white liquid settling below the black liquid if the thickness of the segment of the black liquid is 2 inches, find the radius of the bowl in inches. 𝜋ℎ2 (3𝑟 − ℎ) 𝑉= 3 𝜋(6)2 (3𝑟 − 6) 2𝜋(4)2 (3𝑟 − 4) = 3 3 3(3𝑟 − 6) = 2(16)(3𝑟 − 4) 27𝑟 − 54 = 24𝑟 − 32 𝒓 = 𝟕. 𝟑𝟑𝒊𝒏

Page 19 of 25

d. A sphere having diameter of 30cm is cut into 2 segments. The altitude of the first segment is 6cm. What is the ratio of the area of the second segment to that of the first. 𝐴2 2𝜋𝑟𝐻 24 = = =4 𝐴1 2𝜋𝑟ℎ 6

e. A bowl in the form of a spherical segment with 2 bases has a height of 0.10m. The upper base is a great circle with a diameter of 0.60m. Compute the capacity of the bowl in cu.cm. 4𝜋𝑟 3 𝜋ℎ2 (3𝑟 − ℎ) 𝑉= − 6 3 3 4𝜋(0.3) 𝜋(0.2)2 (3(0.3) − 0.2) − = 𝟎. 𝟏𝟔𝒎𝟑 6 3

Frustum of a Cone 𝝅𝒉(𝑹𝟐 + 𝒓𝟐 + 𝑹𝒓) 𝑽= 𝟑 𝑨𝑳 = 𝝅(𝑹 + 𝒓)𝑳 Example: A frustum of a cone has an upper base radius of 3m and a lower base radius of 6m. If the altitude of the frustum is 9m, compute the volume of the frustum and its lateral area. 𝑉=

𝜋ℎ(𝑅 2 + 𝑟 2 + 𝑅𝑟) 𝜋(9)(62 + 32 + 6(3)) = = 3𝜋(36 + 9 + 18) = 𝟓𝟗𝟑. 𝟕𝟔𝒎𝟑 3 3

𝐿 = √92 + 1.52 = 9.49 𝐴𝐿 = 𝜋(𝑅 + 𝑟)𝐿 = 𝜋(6 + 3)(9.49) = 𝟐𝟔𝟖. 𝟑𝟐𝒎𝟐 Frustum of a Regular Pyramid 𝒉(𝒃 + 𝑩 + √𝑩𝒃) 𝑽= 𝟑 Example: a. The volume of the frustum of a regular triangular pyramid is 135 cu.m. The lower and upper bases are both equilateral triangle with sides equal to 9m and 3m respectively. Find its altitude. 32 𝑆𝑖𝑛60 = 𝟑. 𝟗𝒎𝟐 2 92 𝑆𝑖𝑛60 𝐴𝐵 = = 𝟑𝟓. 𝟎𝟕𝒎𝟐 2 𝐴𝑆 =

Page 20 of 25

ℎ(𝑏 + 𝐵 + √𝐵𝑏) 3 ℎ (3.9 + 35.07 + √35.07(3.9)) 135 = 3 405 ℎ= = 𝟖𝒎 50.66 𝑉=

b. A frustum of a regular pyramid has a upper base of 8m x 10m and a lower base of 10m x 100m and an altitude of 5m. Find the volume of the pyramid. 𝐴𝑈 = 10𝑥8 = 𝟖𝟎𝒎𝟐 𝐴𝐿 = 10𝑥100 = 𝟏𝟎𝟎𝟎𝒎𝟐

𝑉=

5 (80 + 1000 + √80(1000)) ℎ(𝑏 + 𝐵 + √𝐵𝑏) = = 𝟐𝟐𝟕𝟏. 𝟒𝟎𝒎𝟑 3 3

Centroids Straight line 𝒙𝟏 + 𝒙𝟐 𝟐 𝒚𝟏 + 𝒚𝟐 𝒚𝒄 = 𝟐 𝒙𝒄 =

Example: A(3,-7) & B(-4,6) 3 + −4 −𝟏 𝑥𝑐 = = 2 𝟐 𝑦𝑐 =

−7 + 6 −𝟏 = 2 𝟐

Plane Figures Square 𝒃 𝒙𝒄 = 𝟐 𝒉 𝒚𝒄 = 𝟐 Rectangle 𝒃 𝒙𝒄 = 𝟐 𝒉 𝒚𝒄 = 𝟐 Page 21 of 25

Triangle From the right angle 𝒃 𝒙𝒄 = 𝟑 𝒉 𝒚𝒄 = 𝟑

Spandrel

n yc xc 𝟐𝒃 𝒏+𝟐 𝒉 𝒚𝒄 = 𝟑 𝒙𝒄 =

Quarter Circle

yc xc

𝑨=

𝝅𝒓𝟐 𝟒𝒓 𝟒𝒓 , 𝒙𝒄 = , 𝒚𝒄 = 𝟒 𝟑𝝅 𝟑𝝅

Page 22 of 25

Quarter Circle

yc r 𝑨=

𝝅𝒓𝟐 , 𝟒

𝒚𝒄 =

𝟒𝒓 𝟑𝝅

Parabola

b yc a 𝒚𝒄 =

𝟐𝒃 𝟓

𝑨=

𝟐(𝟐𝒂)(𝒃) 𝟑

a

Semi-Ellipse b xc yc b b

a

2a 𝑨=

𝝅𝒂𝒃 𝟐

𝒚𝒄 =

𝟒𝒃 𝟑𝝅

Page 23 of 25

𝒙𝒄 =

𝟒𝒂 𝟑𝝅

𝑨𝑻 ̅ 𝒙 = ∑ 𝑨𝒊 𝒙𝒊 ̅ = ∑ 𝑨𝒊 𝒚 𝑨𝑻 𝒚

Example:

6

5

A3 A1 A2 7

3

𝐴1 = 5𝑥7 = 35𝑠𝑢 𝑥 = 3.5 𝑦 = 2.5 1 𝐴2 = 𝑥3𝑥5 = 7.5𝑠𝑢 2 1 𝑥 = 7 + (3) = 8 3 1 5 𝑦 = (5) = 3 3 1 𝐴3 = 𝑥7𝑥6 = 21𝑠𝑢 2 2 14 𝑥 = (7) = 3 3 1 𝑦 = (6) + 5 = 7 3 Page 24 of 25

𝐴 𝑇 = 35 + 7.5 + 21 = 63.5𝑠𝑢 𝐴 𝑇 𝑥̅ = ∑ 𝐴𝑖 𝑥𝑖 14 63.5𝑥̅ = 35(3.5) + 7.5(8) + 21 ( ) 3 ̅ = 𝟒. 𝟒𝟐𝒖 𝒙 5 63.5𝑦̅ = 35(2.5) + 7.5 ( ) + 21(7) 3 ̅ = 𝟑. 𝟖𝟗𝒖 𝒚

Page 25 of 25